That's not a single 200 W panel, I hope. 1 hp = 746 W, so 5 \$\times\$ 3 hp is good for 11.2 kW, if that 3 hp is the consumed electric power (most likely). If it's delivered mechanical power you'll have to divide by the pump's efficiency, in practice that could be doubling.

If your solar farm can supply the power there's no reason why the pumps should not be connected together.

First, the annual solar irradiance is calculated by taking the power incident on a square meter when perpendicular to the sun, and multiplying by 24 hours and then by 365 days. Since the irradiance is typically about 1.366 kW/sq m, when you multiply it out you get 11.9 kW-hr/sq m, which is your number.

But you might have noticed a problem. This assumes the sun is shining 24 hours per day, which is hardly proper.

Second, your second calculation also assumes 24 hours per day sunlight, which is not right. Since the average length of day is about 12 hours, your 25.6 Wh should be divided by 2, giving 12.8 Wh per day.

Finally, this number tells you how much power is falling on the solar cell, not how much the cell puts out. It also assumes the cell will be perpendicular to the sun at all times, and that the atmosphere does not attenuate sunlight. The first may be true, the second is not. Consider that the sun is much dimmer at sunset than at noon.

Let's take the case of noon sunlight. Assuming the air is very clear and you are getting the nominal irradiance, the power falling on the cell will be 1366 w/sq m x .0078, or about 10 watts. Since you have a solar cell rated for 1 watt, this says that you can assume the cell has an efficiency of about 10%, which is about right.

For a 1-watt solar cell which is fixed in position, and is perpendicular to the sun at noon, you can figure on a total output of 4-6 Wh on a clear day. On the one hand, as the sun moves away from the noon position, the effective area decreases (reaching zero at sunrise and sunset when the cell is edge-on to the sun). Also, sunlight gets attenuated by the atmosphere at lower angles, and finally the cells themselves will typically become less efficient at lower intensities.

## Best Answer

Assuming the current/voltage relationship is linear (it's not, but this gives you a crude lower bound), you could measure the short-circuit current and the open-cell voltage and do

`1/4 * I * V`

to obtain the maximum theoretical power given a worst-case 0.25 fill factor. However a more reasonable value might be obtained by using a different factorBecause the I-V curve is non-linear you either need to find the parameters of your cell and apply equations (such as found here), or simply do it numerically by taking measurements at various I/V and finding the largest

`I*V`

coordinate.